Class-10 (Navier-Stokes Equations)

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Angular deformation

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• Angular deformation of the fluid particle is the sum of the two angular deformations

∆𝜶 + ∆𝜷

• Also ∆𝜶 =∆𝜼

∆𝒙 and ∆𝜷 =

∆𝝃

∆𝒚 , ∆𝜼 𝐚𝐧𝐝 ∆𝝃 are given by ∆𝜼 =

𝝏𝒗

𝝏𝒙∆𝒙∆𝒕 and ∆𝝃 =

𝝏𝒖

𝝏𝒚∆𝒚∆𝒕

• Rate of angular deformation in x-y- plane: lim∆𝒕→𝟎

∆𝜶+∆𝜷 ∆𝒕

= lim∆𝒕→𝟎

∆𝜼

∆𝒙+

∆𝝃

∆𝒚

∆𝒕

• Hence Rate of angular deformation in the x-y- plane: 𝝏𝒖

𝝏𝒚+

𝝏𝒗

𝝏𝒙

• Rate of angular deformation in y-z- plane and x-z- plane: 𝝏𝒗

𝝏𝒛+

𝝏𝒘

𝝏𝒚 and

𝝏𝒘

𝝏𝒙+

𝝏𝒖

𝝏𝒛

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Newtonian and Non-Newtonian Fluids • For Newtonian fluids, the shear stress is linearly proportional to shear strain

Ex: Air and gases, water, kerosene, oil based liquids

• Non-Newtonian fluids, shear stress is not linearly proportional to shear strain.

Ex: Slurries, colloidal suspensions, polymer solutions, blood, paste and cake batter

• Shear thinning and shear thickening fluids:

• Some non-Newtonian fluids become less viscous when more sheared. These are

called shear thinning fluids (Eg. Paint)

• Some non-Newtonian fluids become more viscous when more sheared. These are

called shear thickening or dilatant fluids (Eg. Quick sand and water mixture)

• If shear thinning is extreme, it is called plastic fluid

• If fluid retains the original shape after shear force is

removed, it is called visco-elastic liquids

The angular velocity about the z axis is

Similarly,

• A fluid at rest will have the following stress tensor (there is no shear stress)

• Where P is the hydrostatic pressure , and is equal to the thermodynamic pressure which depends on the type of gas, temperature and density

• When the fluid is in motion, pressure is there and

shear stresses are acting due to viscosity

The new stress tensor is called the viscous stress tensor

Or the deviatoric stress tensor 5/8/2012 4

Stresses on a fluid – at rest and in motion

Shear Stress and Viscosity – 2 dimensional flow

• The force is applied in the y-plane and in the x- direction. Hence shear stress

𝝉𝒚𝒙 =𝜹𝑭𝒙

𝜹𝑨𝒚=

𝒅𝑭𝒙

𝒅𝑨𝒚 , the deformation rate = lim

𝛿𝒕→𝟎

𝜹𝜶 𝜹𝒕

=𝒅𝜶

𝒅𝒕

𝜹𝒍 = 𝜹𝒖𝜹𝒕 also for small angles 𝜹𝒍 = 𝜹𝒚𝜹𝜶

SO, 𝜹𝜶 𝜹𝒕

= 𝜹𝒖 𝜹𝒚

, taking limits on both sides, 𝒅𝜶 𝒅𝒕

= 𝒅𝒖 𝒅𝒚

• Hence the shear rate of the fluid is equal to 𝒅𝒖 𝒅𝒚

• For a Newtonian fluid the shear stress is linearly related to the shear rate

𝝉𝒚𝒙 ∝ 𝒅𝜶 𝒅𝒕

= 𝒅𝒖 𝒅𝒚

• And then 𝝉𝒚𝒙 = 𝛍𝒅𝒖 𝒅𝒚

𝛍 is the viscosity of the fluid

For a 3-dimensional flow, the shear stress becomes complicated, and 𝝉𝒚𝒙 = 𝛍(𝒅𝒖 𝒅𝒚

+𝒅𝒗 𝒅𝒙

) 5/8/2012

5

• In an incompressible fluid, density is constant, and then the thermodynamic pressure becomes the mechanical pressure P

• It has been proved that

• Also, for a Newtonian fluid, the shear (viscous) stress is linearly related to the shear strain, and then where is the stress tensor and the strain tensor

• Hence the viscous stress tensor becomes

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Stresses on a fluid – Incompressible fluid

• Hence the total stress tensor becomes,

• 𝜎𝑥𝑥 = −𝑝 + 2𝜇𝜕𝑢

𝜕𝑥 𝜎𝑦𝑦 = −𝑝 + 2𝜇

𝜕𝑣

𝜕𝑦 𝜎𝑧𝑧 = −𝑝 + 2𝜇

𝜕𝑤

𝜕𝑧

• But 𝑝 = −1

3(𝜎𝑥𝑥 + 𝜎𝑦𝑦 + 𝜎𝑧𝑧)

• So 𝜎𝑥𝑥 =1

3(−𝑝 + 2𝜇

𝜕𝑢

𝜕𝑥 + −𝑝 + 2𝜇

𝜕𝑣

𝜕𝑦 + −𝑝 + 2𝜇

𝜕𝑤

𝜕𝑧) + 2𝜇

𝜕𝑢

𝜕𝑥= −𝑝 −

2

3𝜇∇. 𝐕 + 2𝜇

𝜕𝑢

𝜕𝑥

• 𝜎𝑦𝑦 = −𝑝 −2

3𝜇𝛻. 𝐕 + 2𝜇

𝜕𝑣

𝜕𝑦 and 𝜎𝑧𝑧 = −𝑝 −

2

3𝜇𝛻. 𝐕 + 2𝜇

𝜕𝑤

𝜕𝑧

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Stresses on a fluid – Incompressible fluid

The differential momentum equation

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The differential momentum equation is

𝝆𝒈𝒙 +𝝏𝝈𝒙𝒙

𝝏𝒙+

𝝏𝝉𝒚𝒙

𝝏𝒚+

𝝏𝝉𝒛𝒙

𝝏𝒛= 𝝆(

𝝏𝒖

𝝏𝒕+ 𝒖

𝝏𝒖

𝝏𝒙+v

𝝏𝒖

𝝏𝒙+

𝝏𝒖

𝝏𝒙)

𝝆𝒈𝒚 +𝝏𝝉𝒙𝒚

𝝏𝒙+

𝝏𝝈𝒚𝒚

𝝏𝒚+

𝝏𝝉𝒛𝒚

𝝏𝒛= 𝝆(

𝝏𝒗

𝝏𝒕+ 𝒖

𝝏𝒗

𝝏𝒙+v

𝝏𝒗

𝝏𝒚+

𝝏𝒗

𝝏𝒛)

𝝆𝒈𝒛 +𝝏𝝈𝒙𝒛

𝝏𝒙+

𝝏𝝉𝒚𝒛

𝝏𝒚+

𝝏𝝈𝒛𝒛

𝝏𝒛= 𝝆(

𝝏𝒘

𝝏𝒕+ 𝒖

𝝏𝒘

𝝏𝒙+v

𝝏𝒘

𝝏𝒚+

𝝏𝒘

𝝏𝒛)

• Where p is the Mechanical pressure (incompressible flow)

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The Navier-Stokes Equations • Substituting the value of the stresses in the differential momentum equations,

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• For an incompressible flow,

For incompressible flow, 𝛁. 𝐕 = 𝟎

𝝆𝑫𝒖

𝑫𝒕= 𝝆𝒈𝒙 −

𝝏𝒑

𝝏𝒙+ 𝝁(

𝒅𝟐𝒖

𝒅𝒙𝟐 +𝒅𝟐𝒖

𝒅𝒚𝟐 +𝒅𝟐𝒖

𝒅𝒛𝟐) + 𝝁(𝒅𝟐𝒖

𝒅𝒙𝟐 +𝒅𝟐𝒖

𝒅𝒚𝒅𝒙+

𝒅𝟐𝒖

𝒅𝒛𝒅𝒙)

= 𝝆𝒈𝒙 −𝝏𝒑

𝝏𝒙+ 𝝁(

𝒅𝟐𝒖

𝒅𝒙𝟐 +𝒅𝟐𝒖

𝒅𝒚𝟐 +𝒅𝟐𝒖

𝒅𝒛𝟐 ) + 𝝁𝒅

𝒅𝒙(

𝒅𝒖

𝒅𝒙+

𝒅𝒗

𝒅𝒚+

𝒅𝒘

𝒅𝒛)

= 𝝆𝒈𝒙 −𝝏𝒑

𝝏𝒙+ 𝝁(

𝒅𝟐𝒖

𝒅𝒙𝟐 +𝒅𝟐𝒖

𝒅𝒚𝟐 +𝒅𝟐𝒖

𝒅𝒛𝟐 )

Similarly

𝝆𝑫𝒗

𝑫𝒕 = 𝝆𝒈𝒚 −

𝝏𝒑

𝝏𝒚+ 𝝁(

𝒅𝟐𝒗

𝒅𝒙𝟐 +𝒅𝟐𝒗

𝒅𝒚𝟐 +𝒅𝟐𝒗

𝒅𝒛𝟐)

𝝆𝑫𝒘

𝑫𝒕 = 𝝆𝒈𝒛 −

𝝏𝒑

𝝏𝒛+ 𝝁(

𝒅𝟐𝒘

𝒅𝒙𝟐 +𝒅𝟐𝒘

𝒅𝒚𝟐 +𝒅𝟐𝒘

𝒅𝒛𝟐 )

The Navier-Stokes Equations